Question

Differentiate the following w.r.t. x : (logx)^x – (cos x)^cotx 

Answer 1

Let y = (log x)^x – (cos x)^cotx 
Put u = (log x)^x and v = (cos x)^cotx 
Then y = u – v
∴ `"dy"/"dx" = "du"/"dx" - "dv"/"dx"`  ...(1)
Take u = (log x)^x
∴ log u = log(log x)^x = x.log(log x)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[x.log(logx)]`

= `x"d"/"dx"[log(logx)] + log(logx)."d"/"dx"(x)`

= `x xx 1/logx."d"/"dx"(logx) + log(logx) xx 1`

= `x xx 1/logx xx 1/x + log(logx)`

∴ `"du"/"dx" = u[1/logx + log(logx)]`

= `(logx)^x[1/logx + log(logx)]`   ...(2)
Also v = (cos x)^cotx
∴ log v = log(cos x)^cotx = (cot x).(log cos x)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/dx"[(cotx).log(cosx)]`

= `(cotx)."d"/"dx"(log cosx) + (log cosx)."d"/"dx"(cotx)`

= `cotx xx 1/cosx."d"/"dx"(cosx) + (logcosx)(-"cosec"^2x)`

= `cotx xx 1/cosx xx (-sinx) - ("cosec"^2x)(logcosx)`

∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(logcosx)]`

= –(cos x)^cotx[1 + (cosec²x)(log cos x)] ....(3)
From (1), (2) and (3), we get

∴ `"dy"/"dx" = (logx)^x[1/logx + log(logx)] + (cosx)^cotx[1 + ("cosec"^2x)(logcosx)]`.